13 Solutions for Selected Exercises
253
Exercise 10.2
Inserting the definition of the momentum operator into the Hamiltonian, we find
H = −
2
2m
∂
2
∂ x 2 +
mω
2
2
x
2
.
(13.60)
We therefore need to calculate the derivatives of the wave function ψ(x) with respect
to x and find
ψ
(x) = −β
2 xψ(x)
and
ψ
(x) =
β
4 x
2
− β
2
ψ(x) .
(13.61)
For the matrix element ψ|H |ψ, requested in (a), we then find
ψ|H |ψ =
∞
−∞
−
2
2m
β
4 x
2
− β
2
+
mω
2
2
x
2
ψ(x)
2 dx
= −
2
2m
β
4 1
2β 2 − β
2
+
mω
2
2
1
2β 2 =
ω
4
+
ω
4
, (13.62)
where we used
x
2
ψ
2
(x)dx = 1/2β
2 in the second equality. Incidentally, the first
term in the last equality equals the expectation value of the kinetic energy, requested
in part (c). The second term equals that of the potential energy, requested in (d). The
expectation value of the position operator, requested in (b), is given by ψ|x|ψ ∝
xψ
2 dx = 0. Likewise, we find the expectation value of the momentum operator
from (c) to be zero.
Exercise 10.3
Inserting the substitution x = z + ˆ
rt and x
= z
, mentioned below (5.11), causes
(5.11) to have the same form as (10.22), only the symbols for x and z are exchanged.
Exercise 10.4
The MATLAB code ex10_4.m from the ESM illustrates how to repeatedly calculate
the integral for increasing number N of random numbers. It turns out that we need
around N random numbers for the integral I to stabilize at the I /I ≈ 1/
√
N level.
Thus about N = 10
6 samples are required to reach the 10
−3 level.
Exercise 10.5
The following code snippet first defines the Cauchy distribution cauchy() from
(9.4) with a = 1 as an anonymous function. After initializing the starting value x 0
and the β-parameter for the Metropolis-Hastings algorithm, it passes the function
cauchy() to the metropolis routine from Appendix B.5. Finally, the histogram
of the generated random numbers is shown with the function cauchy(), suitably
scaled, superimposed.
253
Exercise 10.2
Inserting the definition of the momentum operator into the Hamiltonian, we find
H = −
2
2m
∂
2
∂ x 2 +
mω
2
2
x
2
.
(13.60)
We therefore need to calculate the derivatives of the wave function ψ(x) with respect
to x and find
ψ
(x) = −β
2 xψ(x)
and
ψ
(x) =
β
4 x
2
− β
2
ψ(x) .
(13.61)
For the matrix element ψ|H |ψ, requested in (a), we then find
ψ|H |ψ =
∞
−∞
−
2
2m
β
4 x
2
− β
2
+
mω
2
2
x
2
ψ(x)
2 dx
= −
2
2m
β
4 1
2β 2 − β
2
+
mω
2
2
1
2β 2 =
ω
4
+
ω
4
, (13.62)
where we used
x
2
ψ
2
(x)dx = 1/2β
2 in the second equality. Incidentally, the first
term in the last equality equals the expectation value of the kinetic energy, requested
in part (c). The second term equals that of the potential energy, requested in (d). The
expectation value of the position operator, requested in (b), is given by ψ|x|ψ ∝
xψ
2 dx = 0. Likewise, we find the expectation value of the momentum operator
from (c) to be zero.
Exercise 10.3
Inserting the substitution x = z + ˆ
rt and x
= z
, mentioned below (5.11), causes
(5.11) to have the same form as (10.22), only the symbols for x and z are exchanged.
Exercise 10.4
The MATLAB code ex10_4.m from the ESM illustrates how to repeatedly calculate
the integral for increasing number N of random numbers. It turns out that we need
around N random numbers for the integral I to stabilize at the I /I ≈ 1/
√
N level.
Thus about N = 10
6 samples are required to reach the 10
−3 level.
Exercise 10.5
The following code snippet first defines the Cauchy distribution cauchy() from
(9.4) with a = 1 as an anonymous function. After initializing the starting value x 0
and the β-parameter for the Metropolis-Hastings algorithm, it passes the function
cauchy() to the metropolis routine from Appendix B.5. Finally, the histogram
of the generated random numbers is shown with the function cauchy(), suitably
scaled, superimposed.